#### Abstract

Let be complete, simply connected n-dimensional Riemannian manifolds without conjugate points. Assume that is starshaped where . For every point , define : y lies on some geodesic segment in S from x to a point of kerS. There is a finite collection of all maximal sets whose union is S. Further, ker in .

#### 1. Introduction

We begin with some definitions from . A subset in a Riemannian manifold is convex if, for each pair point , there is a unique minimal geodesic segment from to and this segment is in . When dealing with a subset , the word “a unique minimal geodesic segment” should be replaced by “the geodesic.” A subset in a Riemannian manifold is starshaped if there is a point such that, for all , there is a unique minimal geodesic segment from to and this segment is in . The subset of consisting of all points like is called the kernel of   . In , a subset is starshaped if there is a point such that, for all , the geodesic segment joining and is contained in . The subset is convex if and only if . Throughout the paper, and will denote the interior and the boundary of the subset , respectively. Let denote the distance between the two points . For the pair of points , will denote the geodesic segment joining , . For and , . The distance to is locally maximal at if there is some neighborhood of such that   . Finally, all manifolds, maps, fields, and so forth are discussions to make sense. All curves are parametrized by arc-length.

Readers may refer to  for discussions concerning visibility via geodesic segment and starshaped sets. The behavior geodesic in has been discussed by many geometrers as Eberlein  and Goto . For more properties of convexity and starshapedness in , see .

#### 2. Main Results

Let be 2-dimensional manifolds without conjugate points, and assume that is starshaped. For every , we define . Clearly .

Lemma 1. Let be starshaped in 2-dimensional manifolds without conjugate point. Then (a)the set is convex,(b)if is any geodesic segment in joining points of , then ,(c)for every point , there is a unique point of closest to ; that is, there is a unique point of such that .

Proof. We have the following.(a)Since is a starshaped subset of , by , is convex.(b)Since is convex, clearly any geodesic segment in joining points of is in .(c)Let be an arbitrary point in . If , then the result is trivial. . Let there be two points and in closest to , that is, , to show that . Suppose on the contrary that . Since and are in , then . Let , such that and . However, is closer to than and , contradicting the assumption that and are closest to ; then .

Proposition 2. Let be starshaped in 2-dimensional manifolds without conjugate points. For every point , one defines . For any pair of points , if , then .

Proof. Let ; then where . Since , then where . This implies that ; then . We conclude that and the proof is complete.

Proposition 3. Let be a starshaped set. Let . If , then there is some point such that and hence . Moreover, is not locally maximal at .

Proof. Let and . Let be a geodesic segment from to where . Then is a geodesic segment in from to to . Hence . Then, by Proposition 2, we have . Moreover, , because . Then , and . Finally, it is clear that, for all points , . Then the distance to is not locally maximal at .

Proposition 4. Let . If is a maximal A set, then is locally maximal at .

Proof. Assume on the contrary that every neighborhood of contains some point for which . By Proposition 3, let . Then where . Then , so . By Proposition 2, . Since is maximal, then which means that where , which contradicts

Proposition 5. Let be a starshaped set. Let such that is locally maximal at . Then (a)the point is a vertex of ,(b)for every point near , , is not locally maximal at ,(c) for any point .

Proof. We have the following.(a)Since is locally maximal at , then and every neighborhood of contains some point such that . By Proposition 3, . Then must be a vertex of .(b)It is direct from (a).(c), because for any . Then .

Theorem 6. Let be a starshaped set. For every point , there is a finite subcollection of all maximal . Then, the following conditions hold:(a),(b)for every point , in if and only if is a maximal set.

Proof. Assume that .(a)For , belongs to at least one set . Either is in or where is in . This implies that ; then .The reverse inclusion is immediate, so the sets are equal.(b). Let be in ; then is a maximal set. To prove that is in , assume on the contrary that is not in . Since , by (a), there is in such that , but is a maximal set; this implies that . Then, is not subset of any set , so A(z) is in .

Lemma 7. Let be a starshaped set. Each point . Then (a)the set is convex,(b)if is any geodesic segment in joining points of , then .

Proof. We have the following.(a)Let ; then , because is convex. Moreover . Let , such that , respectively. We will prove that . Assume on the contrary that is not in ; then there is and . Let be an arbitrary point of such that , but which means that is not in , which is a contradiction.(b)It is clear that any geodesic segment in S joining points of is in , because is convex.

Theorem 8. Let be a starshaped set. Then .

Proof. Clearly . Now, we will prove that . Let ; then there is a point such that does not see via . Since , by Theorem 6, there is in such that ; by Lemma 7, is convex; then ; this implies that ; then and the proof is complete.

#### 3. Conclusion

All results of the present work are valid in Euclidean space as manifolds without conjugate points , but the generalization of these results to , is more difficult and is left as open problem.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgment

The authors are very grateful to the anonymous referees for their valuable suggestions and comments, which helped the authors to improve this paper.